Question

if a^+b^+4=1740 and a×b=24 find the value of a and b

Answer 1

I deduced the question from your comments on another answer. Apologies if I shouldn't have done that or something. The corrected question must be:

If $a^3 + b^3 + 4 = 1740$ and $ab = 24$, find the values of a and b.

Answer:

{a, b} = {2, 12}

Step-by-step explanation:

The expression is symmetrical in a and b, so their values are interchangeable. Hence 2 solutions would exist. At least.

Now, given: ab = 24.

Then, $b = \frac{24}{a}$

Substituting in the main equation, we get:

$a^3 + {\frac{24}{a}}^3 = 1736$ ...1

We shall now, looking at the above expression, attempt to compare it with:

$(a + \frac{24}{a})^3 = a^3 + {\frac{24}{a}}^3 + 3*a*\frac{24}{a}(a + \frac{24}{a})$ ...2

Let $z = a + \frac{24}{a}$

So, from 2, $z^3 - 72z = a^3 + {\frac{24}{a}}^3$

Now, equating that with 1, we get:

$z^3 - 72z = 1736$

So, z^3 - 72z - 1736 = 0.

And, I gotta admit, this is messed up. It's messed up, really. I don't wanna say it, but then you're gonna pry it outta me:

By hit-and-trial...

The only solution is z = 14. (the others are imaginary, which 9th Grade wouldn't demand, AFAIK)

And then, we know $a + \frac{24}{a} = 14$

Cross-multiply a, then solve the quadratic:

$a^2 + 24 = 14a$

$=> a^2 - 14a + 24 = 0$

$=> a^2 - 2a - 12a + 24 = 0$

$=> a = 2 or a = 12$

If a = 2, b = 12; and if a = 12, b = 2.

Therefore, the set of {a, b} = {2, 12}.

Note, for Indian 9th Grade, do not quote your final answer as I have. Write the expanded version.